Answer:
Explanation:
⁶₃Li will have 3 protons and 3 neutrons .
mass of proton in amu = 1.00727 amu
mass of neutron in amu = 1.00866 amu
mass of lithium nucleus in amu = 6.01512 amu
mass defect = 3 ( 1.00727 + 1.00866 ) - 6.01512 amu
= .03267 amu
Binding energy = mass defect in amu x 931 Mev
= 30.41 MeV
binding energy per nucleon
no of nucleon = 3 + 3 = 6
binding energy per nucleon = 30.41 / 6 Mev
= 5.068 MeV .
The binding energy of ⁶₃Li nucleus in MeV per nucleon is 5.26 MeV per nucleon.
We can see that Li has three protons and three neutrons. The total mass of the Li nucleus is obtained from;
3(1.007277) + 3(1.008665) = 3.023 + 3.026 = 6.049 amu
Actual mass of Li - 6 = 6.0151 amu
Mass defect = 6.049 amu - 6.0151 amu = 0.0339 amu
We can obtain the binding energy in MeV as follows;
Binding energy = 0.0339 amu × 931 = 31.56 MeV
The binding energy per nucleon in MeV per nucleon = 31.56 MeV/6
= 5.26 MeV per nucleon
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